Difference between revisions of "1992 AJHSME Problems/Problem 11"
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The total frequency is <math> 50+60+40+60+40=250 </math>, with the blue frequency of <math> 60 </math>. Therefore, the precentage that preferred blue is <math> \frac{60}{250}=\boxed{\text{(B)}\ 24\%} </math>. | The total frequency is <math> 50+60+40+60+40=250 </math>, with the blue frequency of <math> 60 </math>. Therefore, the precentage that preferred blue is <math> \frac{60}{250}=\boxed{\text{(B)}\ 24\%} </math>. | ||
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| + | ==See Also== | ||
| + | {{AJHSME box|year=1992|num-b=10|num-a=12}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:09, 5 July 2013
Problem
The bar graph shows the results of a survey on color preferences. What percent preferred blue?
Solution
The total frequency is
, with the blue frequency of
. Therefore, the precentage that preferred blue is
.
See Also
| 1992 AJHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
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